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View Full Version : Bellows compensation for S-K 300mm f5.5 Tele-Xenar

JimboWalker
27-Jan-2019, 16:01
I just purchased a nice S-K 300mm f5.5 Tele-Xenar for my Pacemaker Graphic 4x5. How would I calculate infinity and bellows factor? When focused on some trees about 3 blocks away, I am getting a measurement from shutter to film plane of only 170mm! Should I focus on the horizon for infinity, then measure the bellows draw and use that as my standard? I have always used the "shutter to to film plane" method and seems close enough for my needs. Thanks!

Dan Fromm
27-Jan-2019, 16:27
Should I focus on the horizon for infinity, then measure the bellows draw and use that as my standard?

Yes.

JimboWalker
27-Jan-2019, 17:21
Thought so. I'm really amazed that this lens can also focus so closely! I have not run any real test, but I focused on my window curtains at about 5 feet with my Pacemaker! Thanks for the response .

Emmanuel BIGLER
28-Jan-2019, 10:49
Hello from France!

If you want to compute your bellows factor, or exposure compensation factor (ECF in short here) with a (simple) mathematical formula, you need, in principle, to know an additional parameter of your lens named: pupillar magnification factor (PM in short).

This PM factor is fixed by design for a given non-zoom telephoto lens i.e. a telephoto with non-moving optical elements and consequently a fixed focal length.
PM=1 for all quasi-symmetrical lenses. PM<1 for telephotos.

Unfortunately, the PM factor was not published in manufacturers's data-sheet for large format lenses in the past.

But as explained in detail below, in most photographic situations of landscape and portrait work, except in close-up at 1:1 image magnification ratio, you can safely ignore the strange behavior of the Tele Xenar.

----------------------------------------

Now more details about the pupillar magnification factor and additional exposure corrections.

I have estimated the PM factor for my 360 mm Schneider Tele-Arton to be PM ~= 0.57.

In order to do this estimation, you shoud look at the iris from the front of the lens and then from the back of the lens. In a telephoto, the image of the iris appears smaller when seen from the back. For the 360 tele-Arton the image of the iris is 0.57 times smaller seen from the back (the exit pupil) than seen from he front (the entrance pupil).

The PM factor can also be estimated knowing the focal length and estimating the position of the entrance pupil and the exit pupil with respect to the lens barrel and the position or the focal planes.

Although if I do not know what happens in reality for the 300 mm Tele Xenar, it is likely that the PM factor for the Tele Xenar 300 mm 5.5 is somewhere between 0.5 and 0.7, just to get an idea of a very rough estimate.

The general formulae for the exposure compensation factor (ECF) read as

ECF = ((M+PM)/PM)2 = ( 1 + M/PM)2

where M is the (image)/(object) magnification factor. M=0 at infinity, M=1 at 1:1 ratio.

For quasi-symmetrical lens formulae like many large format lenses (wide-angle or standard lenses) PM is close to unity. Hence when PM=1 we get the usual formula for the ECF:

ECF = ( 1 + M)2

expressed in terms of f-stop full-clicks we get

f-stops = log2(ECF) = 3.32 log10(ECF)

e.g. if ECF = 2, open the lens by one full f-stop.

For any lens formula, it happens that the image magnification factor M is given by the same expression, independant of the PM factor.
Let EXT be the additional bellows extension beyond the focal point needed to get a sharp image of the subjet, we have

EXT = M.f
M = EXT/f

Hence in order to estimate your image magnification factor M, you can easily measure the additional bellows extension beyond the infinity setting needed to properly focus.
With a lens of focal length 300 mm, be it quasi-symmetrical or asymmetrical like a telephoto, M = EXT(in mm)/ 300 mm.
For example if EXT = 100 mm or 4", your image magnification factor with the 300 mm Tele Xenar is 100/300 = 1/3 = 0.33

For a 300 mm quasi-symmetrical lens, the shutter-to image distance @M=0.33 is about 300 + 100 = 400 mm.
For the 300 mm Tele Xenar with a flange-focal distance of 170 mm, @M=0.33 the distance between the shutter and the sharp image is 170+100 = 270 mm.

With a quasi-symmetrical lens formula @M=0.33, the ECF would be
ECF = (1+.33)2 = 1.78 this corresponds to 0.8 f-stops, open between 1/2 f-stop and 1 f-stop.
Say one f-stop because in black and white negative photography it does not hurt to over-expose by 1/2 f-stop ;)

Now assume that you have a telephoto with PM=0.5
PM=0.5 yields a bigger exposure correction than for PM=1.
We have @M=0.33 : ECF = (1+ .33 / .5)2 = (1+0.66)2 = 2.76.
2.76 in terms of f-clicks is about 1.5 f-stop. i.e. about 0.5 f-stop more than expected for a quasi-symmetrical lens formula.

SUMMARY

Even if the PM factor for the 300 mm tele-xenar was about 0.5, the additional correction factor is not huge, about 0.5 f-stop for an image magnification ratio of about 0.33

And for landscape and portrait use, when M is smaller than 0.33, this additional correction (with respect to the regular correction valid for quasi-symmetrical lenses) being smaller than 1/2 f-stop can safely be neglected.

Hence you can take benefit of Mr. Salzgeber's "quick disc" device ...

http://www.salzgeber.at/disc/

... to estimate your image magnification factor M, and directly read the ECF without any maths, from the associated scale.
The "quick disc" scale assumes a quasi-symetrical lens formula, i.e. ECF = (1+M)2.
You can, except if you work at 1:1 ratio, forget about the additional correction needed to take into account the asymmetry of the Tele Xenar.

---

I have attached a diagram showing what happens if you insist on using a telephoto for macro work.
Yes the additional exposure corrections with respect to a symmetrical lens can be huge, but who would insist on using a telephoto for macro work ? ;)

rdeloe
28-Jan-2019, 11:10
Some nice advice from Tim Layton here: https://www.timlaytonfineart.com/blog/2016/10/darkroom-daily-digest-how-i-calculate-bellows-factors-magnification-ratios

And various tips and tricks here: https://www.largeformatphotography.info/bellows-factor.html

I use an app on my Android phone.

Dan Fromm
28-Jan-2019, 11:22
Rob, Emmanuel nailed it.

Tim and the article you linked to presented the most common case, for lenses with pupillary magnification = 1. They're both wrong for telephoto lenses (not that uncommon in LF) and inverted telephoto or retrofocus lenses (rare in LF).

This is a common mistake that is nearly everywhere and, alas, in most macro/closeup books. Lefkowitz (see the list) is the big exception there. Bellow factor is very important for macro work when TTL metering is impossible. Another argument for the Horseman Optical Exposure Computer. See the list for an article about it.

rdeloe
28-Jan-2019, 12:28
That's an important clarification Dan. I use flange focal length where Tim refers to focal length in his formula. I shouldn't have assumed that that was understood when I sent the OP to that post. My bad.

Rob, Emmanuel nailed it.

Tim and the article you linked to presented the most common case, for lenses with pupillary magnification = 1. They're both wrong for telephoto lenses (not that uncommon in LF) and inverted telephoto or retrofocus lenses (rare in LF).

This is a common mistake that is nearly everywhere and, alas, in most macro/closeup books. Lefkowitz (see the list) is the big exception there. Bellow factor is very important for macro work when TTL metering is impossible. Another argument for the Horseman Optical Exposure Computer. See the list for an article about it.

Dan Fromm
28-Jan-2019, 13:36
That's an important clarification Dan. I use flange focal length where Tim refers to focal length in his formula. I shouldn't have assumed that that was understood when I sent the OP to that post. My bad.

Rob, that's a problem but we can't see it in this discussion. And for lenses of normal construction (not telephoto, not retrofocus) in shutter flange-focal distance is usually close to focal length. Not always the case with lenses in barrel.

What's wrong in the links you posted is that they ignore pupillary magnification.

rdeloe
28-Jan-2019, 13:57
One of the links I posted is to the LFF "front page" material on bellows extension. I just had another look and it doesn't discuss pupillary magnification. I don't know how material is added to those, but Emmanuel has gone to some lengths to provide a nice explanation of pupillary magnification (a topic that, to be honest, was news to me!) Who adds material to those pages?

Jac@stafford.net
28-Jan-2019, 14:25
Perhaps one of our experts in optics can answer: a telephoto lens with a positive bellows extension change - does the nodal point change, and if it does, then in what direction does it move?

Dan Fromm
28-Jan-2019, 14:31
One of the links I posted is to the LFF "front page" material on bellows extension. I just had another look and it doesn't discuss pupillary magnification. I don't know how material is added to those, but Emmanuel has gone to some lengths to provide a nice explanation of pupillary magnification (a topic that, to be honest, was news to me!) Who adds material to those pages?

Volunteers write articles, offer them to the moderators or to QT, to whom all credit for this site is due, and sometimes they're posted.

Oh, and by the way, Emmanuel has posted several useful and enlightening comments in discussions here. There are links to some of them, including to post #4 in this discussion, in the list.

After discussion with the mods, I and they agreed to post a link to the list in a sticky in the lens section. I've kept control of the list. This leaves me free to update it without going through the mods and perhaps QT and saves the site from the risk that I might clobber a useful article on it.

IMO, and without naming names, some of the articles here are weak and should probably be revised or removed.

Dan Fromm
28-Jan-2019, 14:32
Perhaps one of our experts in optics can answer: a telephoto lens with a bellows extension change - does the nodal point change, and if it does, then in what direction does it move?

Unit-focusing lens? Everything is fixed. If inter-element spacing is fixed, why should the principal planes move when the lens is focused?

Emmanuel BIGLER
28-Jan-2019, 16:07
Hello Again!

I have prepared another diagram more useful for telephoto aficionados with magnification factors between 0 (= object at infinity) and 0.4 = 1/2.5 which corresponds to a tight portrait in 4x5", with a frame size of 10x12.5" (4x2.5 by 5x2.5).

It can be seen that if you do not exceed a magnification factor M = 0.3, the difference in exposure values between a symmetrical lens (PM=1) and an extreme telephoto (PM=0.5) never exceeds about 0.5 f-stop.
This could easily explain why most authors have ignored the general formulae for bellows factors in asymmetrical lenses.
And since on the Internet, many authors simply cut-paste what they find digitally available, no surprise that those exotic formulae found in some rare printed textbooks of the last century, are largely ignored ;)

---------

Regarding "the nodal point", there are several technical points here, sources of frequent confusions.
Strictly speaking, nodal points N and N' have a very precise meaning in geometrical optics. And they go by pairs, like scissors ;)
And they never move for a lens with fixed elements. Of course they do move in a zoom lens.
For a thick compound lens operating in air, N and N' are identical to principal planes H and H'.
H'=N' is located one focal length in front of the image focal point F'. H=N is located one focal length behind the object focal point F.

For a single thin lens element, H=N=H'=N' are located at the center of the thin lens element. Hence the focal length is easy to estimate by measuring the distance between the single lens element and the focal plane.
For a quasi-symmetrical LF lens, H=N are located close to the shutter, as well as H'=N'. The inter-nodal distance HH' is usually a very small fraction of the focal length e.g. 4% for the Apo Ronar. So for those lenses the flange-focal distance is very close to the focal length.

With asymmetrical lenses, we have to forget about this and it is a bit difficult to understand what is still valid and what is non longer valid.
We still have H'=N' and H=N since we operate in air, and the focal length is defined as usual f=FH = F'H'.
Note that when any compound lens operating ins air is reversed, even if it is asymmetrical, the focal length of the reversed lens does not change.
But H=N and H'=N' in a telephoto can now be located far from each other and far from the shutter / lens board.

The consequence for the 300 mm Tele Xenar with its 170 mm of flange-focal distance, is that N'=H' is floating in air in front of the lens, about 130 (300 - 170) mm in front of the shutter!
If we focus the lens on a fixed, far-distant object, with the rear standard of the camera kept fixed after focusing, and if we rotate the lens around N', the image will not move. This property of N' is valid only for far-distant objects, and is used in panoramic cameras where the lens rotates in front of the film held on a drum.
This experiment could be attempted with a view camera, rotating the lens using the vertical swing movement, for example.
Except that we hardly ever mount a 300 mm Tele Xenar on a 130 mm deep recessed board, so that the front swing axis is located close to N'!

Now if we are speaking about the zero-parallax rotating point, used for panoramic stitching, where we rotate the whole camera for a succession of shots to be stitched, this point is the entrance pupil of the lens.
In a quasi-symmetrical lens, the entrance pupil is located very close to point N=H (not N'=H').
In the 360 Tele Arton, the entrance pupil is located quite far from N and N', but for some specific design reasons unknown to me, is located very close to the shutter, between the shutter and the last lens vertex.
I would expect something very similar for the 300 mm Tele Xenar.

See the 2nd diagram for a summary of what I have found for the 360 Tele Arton, with positions of F, F', N=H, N'=H' and the entrance and exit pupils.

JimboWalker
29-Jan-2019, 16:44
My Goodness!! I did not know my question would start such a conversation! In my mind, to make things simple in the field, I will focus on the horizon, measure my bellows extension from the lens board to the film plane and make note. This measurement I will use as my "infinity". Anything I focus closer on, I will calculate the exposure factor. Does that make sense? Thanks to ALL!!

Dan Fromm
29-Jan-2019, 17:30
Yes. But to get the right answer, you'll have to know the lens' pupillary magnification.

Pere Casals
29-Jan-2019, 19:31
So from the graph, I guess that if we want to focus a near flower and a distant mountain at the same time, by using a tilt, it would happen that a tele would darken more the flower and a retrofocus less, requiring perhaps a different ND graded filter if exposing a Velvia slide...

Dan Fromm
29-Jan-2019, 19:34
No. Magnification is magnification is magnification.

Pere Casals
30-Jan-2019, 03:58
Dan, we have different magnifications in that case, the near flower has a very differerent magnification than the moutain, also because of the tilt the effective bellows extensions are different, so each subject has a different compensation...

Doremus Scudder
30-Jan-2019, 12:46
Yes. But to get the right answer, you'll have to know the lens' pupillary magnification.

Dan,

A question: Why can't one find a bellows extension factor for a telephoto lens by simply adding the difference between the bellows extension at infinity and the focal length to the actual bellows draw and using the formula for non-telephoto ("normal") lenses?

E.g., if I have a 400mm tele that focuses at infinity with, say (for simplicity's sake) 200mm, that would make the difference between focal length and infinity focus 200mm. Let's same I'm focused on something close that would need an extension factor. Why can't I just measure the extension (say it's 400mm) add the 200mm (to make 600mm) and calculate my extension factor as I would for a 400mm lens with 600mm of bellows extension?

Would that at least get me in the ballpark?

Best,

Doremus

Pere Casals
30-Jan-2019, 13:47
by simply adding the difference between the bellows extension at infinity and the focal length to the actual bellows draw

Doremus, let me say my view, I guess that Dan is to agree with next: this doesn't work, the usual formula works for lenses having pupilar magnification close to 1, so a calculation from the equivalent situation is flawed for a lens that has a Pupilar Magnification of 0.57 like the tele arton.

See the Emmanuel's graph, it shows exposure compensation from magnification is different from different lenses, so for a given magnification (say 0.4) a regular plasmat (PM = 1) would have 1 stop conpensation while the tele arton has 1+1/2.

Problem is that the compensation formula is good for quasy-symetric lenses, but it has a bias for lenses that have a pupilar magnification not close to 1, because the compensation factor depends on Pupilar Magnification of the lens, and the general compensation formula is only accurate for PM = 1, so good for many lenses but not for all.

Emmanuel teached us a Lesson of the that kind that is hard to find...

JimboWalker
30-Jan-2019, 15:39
Ok...I just calculated the my PM for my 300mm f5.5 Tele-Xenar is .7. I will never use this lens for anything closer than a Head Shot portrait. Once I find out my bellows draw at infinity, What is the fastest way to figure exposure compensation with my PM and the bellows draw measurement for something like a portrait? I suspect I will have very little time to do a lot of cyphering while actually shooting in the field.

Jac@stafford.net
30-Jan-2019, 15:51
Bracket a few exposures on a test subject. Choose the best. Repeat with selected exposure on important subject. Be happy.

Pere Casals
30-Jan-2019, 15:58
About 1/6 stops more than regular formula says, see posted graph, so imho just the usual compensation

JimboWalker
30-Jan-2019, 16:31
THANKS!!! I can sleep better now! After this God Awful winter calms down, I will get out and shoot!